Poisson Distribution

In Statistics, Poisson distribution is one of the important topics. It is used for calculating the possibilities for an event with the average rate of value. Poisson distribution is a discrete probability distribution. In this article, we are going to discuss the definition, Poisson distribution formula, table, mean and variance, and examples in detail.

Table of Contents:

• Definition
• Formula
• Table
• Mean and variance
• Poisson Distribution Expected Value
• Examples
• FAQs

Poisson Distribution Definition

The Poisson distribution is a discrete probability function that means the variable can only take specific values in a given list of numbers, probably infinite. A Poisson distribution measures how many times an event is likely to occur within “x” period of time. In other words, we can define it as the probability distribution that results from the Poisson experiment. A Poisson experiment is a statistical experiment that classifies the experiment into two categories, such as success or failure. Poisson distribution is a limiting process of the binomial distribution.

A Poisson random variable “x” defines the number of successes in the experiment. This distribution occurs when there are events that do not occur as the outcomes of a definite number of outcomes. Poisson distribution is used under certain conditions. They are:

• The number of trials “n” tends to infinity
• Probability of success “p” tends to zero
• np = 1 is finite

Poisson Distribution Formula

The formula for the Poisson distribution function is given by:

f(x) =(e^{– λ} λ^x)/x!

Where,

e is the base of the logarithm

x is a Poisson random variable

λ is an average rate of value

Poisson Distribution Table

As with the binomial distribution, there is a table that we can use under certain conditions that will make calculating probabilities a little easier when using the Poisson Distribution. The table is showing the values of f(x) = P(X ≥ x), where X has a Poisson distribution with parameter λ. Refer the values from the table and substitute it in the Poisson distribution formula to get the probability value. The table displays the values of the Poisson distribution.

Poisson Distribution Mean and Variance

Assume that, we conduct a Poisson experiment, in which the average number of successes within a given range is taken as λ. In Poisson distribution, the mean of the distribution is represented by λ and e is constant, which is approximately equal to 2.71828. Then, the Poisson probability is:

P(x, λ ) =(e^{– λ} λ^x)/x!

In Poisson distribution, the mean is represented as E(X) = λ.

For a Poisson Distribution, the mean and the variance are equal. It means that E(X) = V(X)

Where,

V(X) is the variance.

Poisson Distribution Expected Value

A random variable is said to have a Poisson distribution with the parameter λ, where “λ” is considered as an expected value of the Poisson distribution.

The expected value of the Poisson distribution is given as follows:

E(x) = μ = d(e^λ(t-1))/dt, at t=1.

E(x) = λ

Therefore, the expected value (mean) and the variance of the Poisson distribution is equal to λ.

Poisson Distribution Examples

An example to find the probability using the Poisson distribution is given below:

Example 1:

A random variable X has a Poisson distribution with parameter λ such that P (X = 1) = (0.2) P (X = 2). Find P (X = 0).

Solution:

For the Poisson distribution, the probability function is defined as:

P (X =x) = (e^{– λ} λ^x)/x!, where λ is a parameter.

Given that, P (x = 1) = (0.2) P (X = 2)

(e^{– λ} λ¹)/1! = (0.2)(e^{– λ} λ²)/2!

⇒λ = λ²/ 10

⇒λ = 10

Now, substitute λ = 10, in the formula, we get:

P (X =0 ) = (e^{– λ} λ⁰)/0!

P (X =0) = e^-10 = 0.0000454

Thus, P (X= 0) = 0.0000454

Example 2:

Telephone calls arrive at an exchange according to the Poisson process at a rate λ= 2/min. Calculate the probability that exactly two calls will be received during each of the first 5 minutes of the hour.

Solution:

Assume that “N” be the number of calls received during a 1 minute period.

Therefore,

P(N= 2) = (e^-2. 2²)/2!

P(N=2) = 2e^-2.

Now, “M” be the number of minutes among 5 minutes considered, during which exactly 2 calls will be received. Thus “M” follows a binomial distribution with parameters n=5 and p= 2e^-2.

P(M=5) = 32 x e^-10

P(M =5) = 0.00145, where “e” is a constant, which is approximately equal to 2.718. 

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